\displaystyle \textbf{Question 1: }\text{Find the perimeter and area of a rectangle whose length and breadth are }20\text{ cm}
\displaystyle \text{and }8\text{ cm respectively.}
\displaystyle \text{Answer:}
\displaystyle \text{Length }(l)=20\text{ cm and Breadth }(b)=8\text{ cm.}
\displaystyle \text{Perimeter}=2(l+b)=2(20+8)=56\text{ cm.}
\displaystyle \text{Area}=l\times b=20\times8=160\text{ cm}^2.
\displaystyle \therefore \text{The perimeter is }56\text{ cm and the area is }160\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{A rectangular room floor is }192\text{ m}^2\text{ in area. If its length is }16\text{ m,}
\displaystyle \text{find its perimeter.}
\displaystyle \text{Answer:}
\displaystyle \text{Length }(l)=16\text{ m. Let the breadth }(b)=x\text{ m.}
\displaystyle \text{Area}=l\times b
\displaystyle 192=16\times x
\displaystyle x=12\text{ m.}
\displaystyle \therefore \text{Breadth}=12\text{ m.}
\displaystyle \text{Perimeter}=2(l+b)=2(16+12)=56\text{ m.}
\displaystyle \therefore \text{The perimeter of the room is }56\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the length of a diagonal of a rectangle whose adjacent sides are }8\text{ m and }6\text{ m long.}
\displaystyle \text{Answer:}
\displaystyle \text{Length }(l)=8\text{ m and Breadth }(b)=6\text{ m.}
\displaystyle \text{Diagonal of the rectangle}=\sqrt{l^2+b^2}
\displaystyle =\sqrt{8^2+6^2}
\displaystyle =\sqrt{64+36}=\sqrt{100}=10\text{ m.}
\displaystyle \therefore \text{The length of the diagonal is }10\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the length of a diagonal of a square of side }4\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Side of the square }(a)=4\text{ cm.}
\displaystyle \text{Diagonal of the square}=\sqrt{2}\times a
\displaystyle =\sqrt{2}\times4=4\sqrt{2}\text{ cm.}
\displaystyle \therefore \text{The length of the diagonal is }4\sqrt{2}\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the perimeter of a square if the sum of the lengths of its diagonals is }144\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the square be }a\text{ cm.}
\displaystyle \text{Let the length of each diagonal be }d\text{ cm.}
\displaystyle \text{Since the sum of the two diagonals is }144\text{ cm,}
\displaystyle 2d=144
\displaystyle d=72\text{ cm.}
\displaystyle \text{Diagonal of a square}=\sqrt{2}\,a
\displaystyle \sqrt{2}\,a=72
\displaystyle a=\frac{72}{\sqrt{2}}=36\sqrt{2}\text{ cm.}
\displaystyle \text{Perimeter of the square}=4a
\displaystyle =4\times36\sqrt{2}=144\sqrt{2}\text{ cm.}
\displaystyle \therefore \text{The perimeter of the square is }144\sqrt{2}\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The length and breadth of a room are in the ratio }3:2.\text{ Its area is }216\text{ m}^2.
\displaystyle \text{Find its perimeter.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length }(l)=3x\text{ m and the breadth }(b)=2x\text{ m.}
\displaystyle \text{Area of the rectangle}=l\times b
\displaystyle 216=3x\times2x
\displaystyle 6x^2=216
\displaystyle x^2=36
\displaystyle x=6.
\displaystyle \therefore l=18\text{ m and }b=12\text{ m.}
\displaystyle \text{Perimeter}=2(l+b)=2(18+12)=60\text{ m.}
\displaystyle \therefore \text{The perimeter of the room is }60\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The diagonal of square }A\text{ is }(a+b).\text{ Find the diagonal of square }B\text{ whose}
\displaystyle \text{area is twice the area of square }A.
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of square }A\text{ be }x.
\displaystyle \text{Diagonal of square }A=\sqrt{2}x=a+b.
\displaystyle \therefore x=\frac{a+b}{\sqrt{2}}.\qquad\ldots\text{(i)}
\displaystyle \text{Let the side of square }B\text{ be }y.
\displaystyle \text{Since the area of square }B\text{ is twice the area of square }A,
\displaystyle y^2=2x^2
\displaystyle \therefore y=\sqrt{2}x.
\displaystyle \text{Diagonal of square }B=\sqrt{2}y
\displaystyle =\sqrt{2}\times\sqrt{2}x=2x.\qquad\ldots\text{(ii)}
\displaystyle \text{Substituting (i) in (ii),}
\displaystyle \text{Diagonal of square }B=2\left(\frac{a+b}{\sqrt{2}}\right)=\sqrt{2}(a+b).
\displaystyle \therefore \text{The diagonal of square }B\text{ is }\sqrt{2}(a+b).
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The perimeter of a square is }(4x+20)\text{ cm. Find its diagonal.}
\displaystyle \text{Answer:}
\displaystyle \text{Perimeter of the square}=4x+20\text{ cm.}
\displaystyle \therefore \text{Side}=\frac{4x+20}{4}=x+5\text{ cm.}
\displaystyle \text{Diagonal of a square}=\sqrt{2}\times\text{Side}
\displaystyle =\sqrt{2}(x+5)\text{ cm.}
\displaystyle \therefore \text{The diagonal of the square is }\sqrt{2}(x+5)\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find the area of a square that can be inscribed in a circle of radius }10\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the circle}=10\text{ cm.}
\displaystyle \therefore \text{Diameter of the circle}=20\text{ cm.}
\displaystyle \text{The diagonal of the inscribed square equals the diameter of the circle.}
\displaystyle \therefore \text{Diagonal of the square}=20\text{ cm.}
\displaystyle \text{Let the side of the square be }a\text{ cm.}
\displaystyle a\sqrt{2}=20
\displaystyle a=\frac{20}{\sqrt{2}}=10\sqrt{2}\text{ cm.}
\displaystyle \text{Area of the square}=a^2
\displaystyle =(10\sqrt{2})^2=200\text{ cm}^2.
\displaystyle \therefore \text{The area of the square is }200\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the perimeter of a square if the sum of the lengths of its diagonals is }100\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the square be }x\text{ cm.}
\displaystyle \text{Diagonal of the square}=\sqrt{2}x\text{ cm.}
\displaystyle \text{Sum of the two diagonals}=100\text{ cm.}
\displaystyle 2(\sqrt{2}x)=100
\displaystyle \sqrt{2}x=50
\displaystyle x=\frac{50}{\sqrt{2}}=25\sqrt{2}\text{ cm.}
\displaystyle \text{Perimeter of the square}=4x
\displaystyle =4\times25\sqrt{2}=100\sqrt{2}\text{ cm.}
\displaystyle \therefore \text{The perimeter of the square is }100\sqrt{2}\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{The diagonal of a square is }14\text{ cm. Find its area.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the square be }x\text{ cm.}
\displaystyle \text{Diagonal of the square}=\sqrt{2}x.
\displaystyle \sqrt{2}x=14
\displaystyle x=\frac{14}{\sqrt{2}}=7\sqrt{2}\text{ cm.}
\displaystyle \text{Area of the square}=x^2
\displaystyle =(7\sqrt{2})^2=98\text{ cm}^2.
\displaystyle \therefore \text{The area of the square is }98\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Find the area and perimeter of a square plot of land whose diagonal is }15\text{ m.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the square be }x\text{ m.}
\displaystyle \text{Diagonal of the square}=\sqrt{2}x.
\displaystyle \sqrt{2}x=15
\displaystyle x=\frac{15}{\sqrt{2}}=\frac{15\sqrt{2}}{2}\text{ m.}
\displaystyle \text{Area of the square}=x^2
\displaystyle =\left(\frac{15}{\sqrt{2}}\right)^2=\frac{225}{2}=112.5\text{ m}^2.
\displaystyle \text{Perimeter of the square}=4x
\displaystyle =4\times\frac{15}{\sqrt{2}}=30\sqrt{2}\text{ m.}
\displaystyle \therefore \text{The area is }112.5\text{ m}^2\text{ and the perimeter is }30\sqrt{2}\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Find the ratio of the area of a square to that of the square drawn on its diagonal.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the square be }x.
\displaystyle \therefore \text{Diagonal}=\sqrt{2}x.
\displaystyle \text{Area of the first square}=x^2.
\displaystyle \text{Area of the square drawn on its diagonal}=(\sqrt{2}x)^2=2x^2.
\displaystyle \therefore \text{Required ratio}=x^2:2x^2=1:2.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The diagonal of square }A\text{ is }(a+b).\text{ Find the diagonal of square }B\text{ whose}
\displaystyle \text{area is half of the area of square }A.
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of square }A\text{ be }x.
\displaystyle \sqrt{2}x=a+b.
\displaystyle \therefore x=\frac{a+b}{\sqrt{2}}.\qquad\ldots\text{(i)}
\displaystyle \text{Let the side of square }B\text{ be }y.
\displaystyle \text{Since the area of square }B\text{ is half the area of square }A,
\displaystyle y^2=\frac{1}{2}x^2
\displaystyle \therefore y=\frac{x}{\sqrt{2}}.
\displaystyle \text{Diagonal of square }B=\sqrt{2}y
\displaystyle =\sqrt{2}\times\frac{x}{\sqrt{2}}=x.
\displaystyle \text{Using (i),}
\displaystyle \text{Diagonal of square }B=\frac{a+b}{\sqrt{2}}.
\displaystyle \therefore \text{The diagonal of square }B\text{ is }\frac{a+b}{\sqrt{2}}.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The perimeter of a square is }48\text{ m. The area of a rectangle is }4\text{ m}^2\text{ less}
\displaystyle \text{than the area of the given square. If the length of the rectangle is }14\text{ m, find its breadth.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the square be }a\text{ m.}
\displaystyle 4a=48
\displaystyle a=12\text{ m.}
\displaystyle \text{Area of the square}=12^2=144\text{ m}^2.
\displaystyle \text{Area of the rectangle}=144-4=140\text{ m}^2.
\displaystyle \text{Let the breadth of the rectangle be }b\text{ m.}
\displaystyle 14\times b=140
\displaystyle b=10\text{ m.}
\displaystyle \therefore \text{The breadth of the rectangle is }10\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The perimeter of one square is }748\text{ cm and that of another is }336\text{ cm. Find the}
\displaystyle \text{perimeter and diagonal of a square whose area is equal to the sum of the areas of these two squares.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the first square be }a_1\text{ cm.}
\displaystyle 4a_1=748
\displaystyle a_1=187\text{ cm.}
\displaystyle \text{Area of the first square}=187^2=34969\text{ cm}^2.
\displaystyle \text{Let the side of the second square be }a_2\text{ cm.}
\displaystyle 4a_2=336
\displaystyle a_2=84\text{ cm.}
\displaystyle \text{Area of the second square}=84^2=7056\text{ cm}^2.
\displaystyle \text{Area of the third square}=34969+7056=42025\text{ cm}^2.
\displaystyle \therefore \text{Side of the third square}=\sqrt{42025}=205\text{ cm.}
\displaystyle \text{Perimeter}=4\times205=820\text{ cm.}
\displaystyle \text{Diagonal}=\sqrt{2}\times205=205\sqrt{2}\text{ cm}
\displaystyle \approx289.91\text{ cm.}
\displaystyle \therefore \text{The perimeter is }820\text{ cm and the diagonal is }205\sqrt{2}\text{ cm}\approx289.91\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The perimeter of a rectangular cardboard is }96\text{ cm. If its breadth is }18\text{ cm,}
\displaystyle \text{find the length and area of the cardboard.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length of the rectangle be }l\text{ cm.}
\displaystyle \text{Breadth }(b)=18\text{ cm.}
\displaystyle 2(l+b)=96
\displaystyle l+18=48
\displaystyle l=30\text{ cm.}
\displaystyle \text{Area}=l\times b=30\times18=540\text{ cm}^2.
\displaystyle \therefore \text{The length of the cardboard is }30\text{ cm and its area is }540\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If the sides of two squares are in the ratio }x:y,\text{ prove that their areas are in}
\displaystyle \text{the ratio }x^2:y^2.
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the first square be }x.
\displaystyle \therefore \text{Area of the first square}=x^2.
\displaystyle \text{Let the side of the second square be }y.
\displaystyle \therefore \text{Area of the second square}=y^2.
\displaystyle \therefore \text{Ratio of their areas}=x^2:y^2.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{In exchange for a square plot one of whose sides is }84\text{ m, a man wants to buy a}
\displaystyle \text{rectangular plot }144\text{ m long and of the same area as the square plot. Find the width of the}
\displaystyle \text{rectangular plot.}
\displaystyle \text{Answer:}
\displaystyle \text{Side of the square plot}=84\text{ m.}
\displaystyle \text{Length of the rectangular plot}=144\text{ m.}
\displaystyle \text{Let the breadth of the rectangular plot be }b\text{ m.}
\displaystyle \text{Area of the square}=\text{Area of the rectangle}
\displaystyle 84\times84=144\times b
\displaystyle b=\frac{84\times84}{144}=49\text{ m.}
\displaystyle \therefore \text{The breadth of the rectangular plot is }49\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{A rectangular lawn }80\text{ m}\times60\text{ m has two roads each }10\text{ m wide running}
\displaystyle \text{through its middle, one parallel to the length and the other parallel to the breadth. Find the}
\displaystyle \text{cost of graveling them at }30\text{ paisa per square metre.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of the road parallel to the length}=80\times10=800\text{ m}^2.
\displaystyle \text{Area of the road parallel to the breadth}=60\times10=600\text{ m}^2.
\displaystyle \text{Area of the overlapping square}=10\times10=100\text{ m}^2.
\displaystyle \therefore \text{Total area of the roads}=800+600-100=1300\text{ m}^2.
\displaystyle \text{Cost of graveling}=1300\times0.30=\text{Rs. }390.
\displaystyle \therefore \text{The cost of graveling the roads is Rs. }390.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{The area of a square plot is }\frac{1}{2}\text{ hectare. Find the diagonal of the square.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the square be }a\text{ m.}
\displaystyle 1\text{ hectare}=10000\text{ m}^2.
\displaystyle \therefore \frac{1}{2}\text{ hectare}=5000\text{ m}^2.
\displaystyle a^2=5000
\displaystyle a=\sqrt{5000}=50\sqrt{2}\text{ m.}
\displaystyle \text{Diagonal of the square}=\sqrt{2}\times a
\displaystyle =\sqrt{2}\times50\sqrt{2}=100\text{ m.}
\displaystyle \therefore \text{The diagonal of the square is }100\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{A lawn is in the form of a rectangle having its sides in the ratio }5:2.\text{ The area of the}
\displaystyle \text{lawn is }1000\text{ m}^2.\text{ Find the cost of fencing it at the rate of Rs. }8.50\text{ per metre.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length}=5x\text{ m and the breadth}=2x\text{ m.}
\displaystyle \text{Area of the rectangle}=1000\text{ m}^2.
\displaystyle 5x\times2x=1000
\displaystyle 10x^2=1000
\displaystyle x=10\text{ m.}
\displaystyle \therefore \text{Length}=50\text{ m and Breadth}=20\text{ m.}
\displaystyle \text{Perimeter}=2(50+20)=140\text{ m.}
\displaystyle \text{Cost of fencing}=140\times8.50=\text{Rs. }1190.
\displaystyle \therefore \text{The cost of fencing the lawn is Rs. }1190.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{The area of a square park is }40,000\text{ m}^2.\text{ Find the cost of fencing it at the rate}
\displaystyle \text{of Rs. }2.80\text{ per metre.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of the square}=40000\text{ m}^2.
\displaystyle \text{Let the side of the square be }a\text{ m.}
\displaystyle a^2=40000
\displaystyle a=200\text{ m.}
\displaystyle \text{Perimeter}=4a=800\text{ m.}
\displaystyle \text{Cost of fencing}=800\times2.80=\text{Rs. }2240.
\displaystyle \therefore \text{The cost of fencing the square park is Rs. }2240.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{The area of the base of a rectangular tank is }2400\text{ m}^2\text{ and its sides are in}
\displaystyle \text{the ratio }3:2.\text{ Find the cost of planting flowers round it at the rate of Rs. }1.25\text{ per metre.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length}=3x\text{ m and the breadth}=2x\text{ m.}
\displaystyle 3x\times2x=2400
\displaystyle 6x^2=2400
\displaystyle x^2=400
\displaystyle x=20\text{ m.}
\displaystyle \therefore \text{Length}=60\text{ m and Breadth}=40\text{ m.}
\displaystyle \text{Perimeter}=2(60+40)=200\text{ m.}
\displaystyle \text{Cost of planting flowers}=200\times1.25=\text{Rs. }250.
\displaystyle \therefore \text{The cost of planting flowers is Rs. }250.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{A rectangular field }242\text{ m long has an area of }4840\text{ m}^2.\text{ What will be the cost}
\displaystyle \text{of fencing the field on all four sides, if }1\text{ metre of fencing costs }20\text{ paisa?}
\displaystyle \text{Answer:}
\displaystyle \text{Length}=242\text{ m and Area}=4840\text{ m}^2.
\displaystyle \text{Breadth}=\frac{4840}{242}=20\text{ m.}
\displaystyle \text{Perimeter}=2(242+20)=524\text{ m.}
\displaystyle 20\text{ paisa}=\text{Rs. }0.20.
\displaystyle \text{Cost of fencing}=524\times0.20=\text{Rs. }104.80.
\displaystyle \therefore \text{The cost of fencing the field is Rs. }104.80.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{A rectangular grassy plot is }112\text{ m}\times78\text{ m. It has a gravel path }2.5\text{ m wide}
\displaystyle \text{all around it on the inside. Find the area of the path and the cost of constructing it at the rate of}
\displaystyle \text{Rs. }3.40\text{ per square metre.}
\displaystyle \text{Answer:}
\displaystyle \text{Outer dimensions of the plot: Length}=112\text{ m, Breadth}=78\text{ m.}
\displaystyle \text{Since the path is }2.5\text{ m wide on each side,}
\displaystyle \text{Inner length}=112-2(2.5)=107\text{ m.}
\displaystyle \text{Inner breadth}=78-2(2.5)=73\text{ m.}
\displaystyle \text{Area of the path}=(112\times78)-(107\times73)
\displaystyle =8736-7811=925\text{ m}^2.
\displaystyle \text{Cost of construction}=925\times3.40=\text{Rs. }3145.00.
\displaystyle \therefore \text{The area of the path is }925\text{ m}^2\text{ and the cost of constructing it is Rs. }3145.00.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{There is a square field whose side is }44\text{ m. A flowerbed is prepared in its centre,}
\displaystyle \text{leaving a gravel path of uniform width all around the flowerbed. The total cost of laying the flowerbed}
\displaystyle \text{and graveling the path at Rs. }2\text{ and Rs. }1\text{ per square metre respectively is Rs. }3536.
\displaystyle \text{Find the width of the gravel path.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the square flowerbed be }x\text{ m.}
\displaystyle \text{Area of the flowerbed}=x^2\text{ m}^2.
\displaystyle \text{Area of the gravel path}=44^2-x^2\text{ m}^2.
\displaystyle \text{Total cost}=2x^2+1(44^2-x^2)
\displaystyle 3536=2x^2+1936-x^2
\displaystyle x^2=3536-1936=1600
\displaystyle x=40\text{ m.}
\displaystyle \text{Total difference in side lengths}=44-40=4\text{ m.}
\displaystyle \text{Width of the path}=\frac{4}{2}=2\text{ m.}
\displaystyle \therefore \text{The width of the gravel path is }2\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{How many tiles of size }40\text{ cm}\times40\text{ cm each are required to pave a}
\displaystyle \text{footpath }1\text{ m wide running all around the outside of a grassy plot }28\text{ m by }18\text{ m?}
\displaystyle \text{Answer:}
\displaystyle \text{Outer length}=28+2(1)=30\text{ m.}
\displaystyle \text{Outer breadth}=18+2(1)=20\text{ m.}
\displaystyle \text{Area of the footpath}=(30\times20)-(28\times18)
\displaystyle =600-504=96\text{ m}^2.
\displaystyle \text{Area of one tile}=40\text{ cm}\times40\text{ cm} \\ =0.4\text{ m}\times0.4\text{ m}=0.16\text{ m}^2.
\displaystyle \text{Number of tiles required}=\frac{96}{0.16}=600.
\displaystyle \therefore \text{600 tiles are required.}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{A room is }7.4\text{ m long, }3.6\text{ m broad and }4\text{ m high. It has two doors}
\displaystyle 2.1\text{ m}\times1.2\text{ m and }5\text{ windows each }1.8\text{ m}\times1.2\text{ m. How much will it cost to}
\displaystyle \text{whitewash the walls of the room at the rate of Rs. }2.00\text{ per square metre?}
\displaystyle \text{Answer:}
\displaystyle \text{Length }(l)=7.4\text{ m, Breadth }(b)=3.6\text{ m and Height }(h)=4\text{ m.}
\displaystyle \text{Area of the four walls}=2(l+b)h
\displaystyle =2(7.4+3.6)\times4=88\text{ m}^2.
\displaystyle \text{Area of two doors}=2\times2.1\times1.2=5.04\text{ m}^2.
\displaystyle \text{Area of five windows}=5\times1.8\times1.2=10.8\text{ m}^2.
\displaystyle \text{Total area of doors and windows}=5.04+10.8=15.84\text{ m}^2.
\displaystyle \text{Area to be whitewashed}=88-15.84=72.16\text{ m}^2.
\displaystyle \text{Cost of whitewashing}=72.16\times2=\text{Rs. }144.32.
\displaystyle \therefore \text{The cost of whitewashing the walls is Rs. }144.32.
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{A carpet is laid on the floor of a room }8\text{ m}\times5\text{ m. There is a border of constant}
\displaystyle \text{width around the carpet. If the area of the border is }1.2\text{ m}^2,\text{ find its width.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the width of the border be }x\text{ m.}
\displaystyle \text{Area of the floor}=8\times5=40\text{ m}^2.
\displaystyle \text{Dimensions of the carpet}=(8-2x)\text{ m}\times(5-2x)\text{ m.}
\displaystyle \text{Area of border}=\text{Area of floor}-\text{Area of carpet}
\displaystyle 1.2=40-(8-2x)(5-2x)
\displaystyle 1.2=40-(40-26x+4x^2)
\displaystyle 1.2=26x-4x^2
\displaystyle 4x^2-26x+1.2=0
\displaystyle 20x^2-130x+6=0
\displaystyle 10x^2-65x+3=0
\displaystyle x=\frac{65\pm\sqrt{65^2-4(10)(3)}}{20}
\displaystyle x=\frac{65\pm\sqrt{4105}}{20}.
\displaystyle \text{Since the width must be less than }2.5\text{ m,}
\displaystyle x=\frac{65-\sqrt{4105}}{20}\approx0.0465\text{ m.}
\displaystyle \therefore \text{The width of the border is approximately }0.0465\text{ m}=4.65\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{A rectangular courtyard, }3.78\text{ m long and }5.25\text{ m broad, is to be paved}
\displaystyle \text{exactly with square tiles, all of the same size. Find the largest size of such a tile and the number}
\displaystyle \text{of tiles required to pave it.}
\displaystyle \text{Answer:}
\displaystyle \text{Length of courtyard}=3.78\text{ m}=378\text{ cm.}
\displaystyle \text{Breadth of courtyard}=5.25\text{ m}=525\text{ cm.}
\displaystyle \text{HCF of }378\text{ and }525=21.
\displaystyle \therefore \text{Side of the largest square tile}=21\text{ cm.}
\displaystyle \text{Number of tiles along the length}=\frac{378}{21}=18.
\displaystyle \text{Number of tiles along the breadth}=\frac{525}{21}=25.
\displaystyle \text{Total number of tiles}=18\times25=450.
\displaystyle \therefore \text{The largest tile is }21\text{ cm}\times21\text{ cm and }450\text{ tiles are required.}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{The cost of fencing a square field at }60\text{ paisa per metre is Rs. }1200.\text{ Find the}
\displaystyle \text{cost of reaping the field at the rate of }50\text{ paisa per }100\text{ m}^2.
\displaystyle \text{Answer:}
\displaystyle 60\text{ paisa}=\text{Rs. }0.60.
\displaystyle \text{Perimeter}\times0.60=1200
\displaystyle \text{Perimeter}=\frac{1200}{0.60}=2000\text{ m.}
\displaystyle \therefore \text{Side of the square}=\frac{2000}{4}=500\text{ m.}
\displaystyle \text{Area of the field}=500\times500=250000\text{ m}^2.
\displaystyle \text{Cost of reaping}=\frac{250000}{100}\times0.50
\displaystyle =2500\times0.50=\text{Rs. }1250.
\displaystyle \therefore \text{The cost of reaping the field is Rs. }1250.
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{A room }4.9\text{ m long and }3.5\text{ m broad is carpeted, leaving an uncovered}
\displaystyle \text{margin }25\text{ cm wide all around the room. If the breadth of the carpet is }80\text{ cm, find its cost}
\displaystyle \text{at Rs. }60\text{ per metre.}
\displaystyle \text{Answer:}
\displaystyle 25\text{ cm}=0.25\text{ m.}
\displaystyle \text{Length of the carpeted portion}=4.9-2(0.25)=4.4\text{ m.}
\displaystyle \text{Breadth of the carpeted portion}=3.5-2(0.25)=3\text{ m.}
\displaystyle \text{Area to be carpeted}=4.4\times3=13.2\text{ m}^2.
\displaystyle \text{Breadth of the carpet}=80\text{ cm}=0.8\text{ m.}
\displaystyle \text{Let the length of carpet required be }l\text{ m.}
\displaystyle l\times0.8=13.2
\displaystyle l=\frac{13.2}{0.8}=16.5\text{ m.}
\displaystyle \text{Cost of carpet}=16.5\times60=\text{Rs. }990.
\displaystyle \therefore \text{The cost of the carpet is Rs. }990.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{The cost of carpeting a room at Rs. }2.50\text{ per square metre is Rs. }450.\text{ The cost}
\displaystyle \text{of whitewashing the walls at }50\text{ paisa per square metre is Rs. }135.\text{ The room is }12\text{ m}
\displaystyle \text{wide. Find its height.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of the floor}=\frac{450}{2.50}=180\text{ m}^2.
\displaystyle \text{Let the length of the room be }l\text{ m.}
\displaystyle 12\times l=180
\displaystyle l=15\text{ m.}
\displaystyle 50\text{ paisa}=\text{Rs. }0.50.
\displaystyle \text{Area of the four walls}=\frac{135}{0.50}=270\text{ m}^2.
\displaystyle \text{Let the height of the room be }h\text{ m.}
\displaystyle 2(l+b)h=270
\displaystyle 2(15+12)h=270
\displaystyle 54h=270
\displaystyle h=5\text{ m.}
\displaystyle \therefore \text{The height of the room is }5\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{A room }5\text{ m long and }4\text{ m wide is surrounded by a verandah. Find the width}
\displaystyle \text{of the verandah if it occupies }22\text{ m}^2.
\displaystyle \text{Answer:}
\displaystyle \text{Let the width of the verandah be }x\text{ m.}
\displaystyle \text{Outer length}=(5+2x)\text{ m and outer breadth}=(4+2x)\text{ m.}
\displaystyle \text{Area of verandah}=(5+2x)(4+2x)-5\times4
\displaystyle 22=(5+2x)(4+2x)-20
\displaystyle 22=20+10x+8x+4x^2-20
\displaystyle 4x^2+18x-22=0
\displaystyle 2x^2+9x-11=0
\displaystyle (2x+11)(x-1)=0
\displaystyle x=1\text{ or }x=-\frac{11}{2}.
\displaystyle \text{Rejecting }x=-\frac{11}{2},\text{ since width cannot be negative, }x=1\text{ m.}
\displaystyle \therefore \text{The width of the verandah is }1\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{A square carpet is spread in the centre of a square room of side }55\text{ dm, leaving}
\displaystyle \text{a margin of equal width all around. The cost of carpeting at }25\text{ paisa per dm}^2\text{ and decorating}
\displaystyle \text{the margin at }15\text{ paisa per dm}^2\text{ is Rs. }703.75.\text{ Find the width of the margin.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the width of the margin be }a\text{ dm.}
\displaystyle \text{Side of the carpet}=(55-2a)\text{ dm.}
\displaystyle \text{Area of the carpet}=(55-2a)^2\text{ dm}^2.
\displaystyle \text{Area of the margin}=55^2-(55-2a)^2\text{ dm}^2.
\displaystyle 25\text{ paisa}=\text{Rs. }0.25\text{ and }15\text{ paisa}=\text{Rs. }0.15.
\displaystyle 0.25(55-2a)^2+0.15\{55^2-(55-2a)^2\}=703.75
\displaystyle 0.10(55-2a)^2+453.75=703.75
\displaystyle (55-2a)^2=2500
\displaystyle 55-2a=50
\displaystyle 2a=5
\displaystyle a=2.5\text{ dm.}
\displaystyle \therefore \text{The width of the margin is }2.5\text{ dm.}
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{If the length and breadth of a room are increased by }1\text{ m, the area is increased}
\displaystyle \text{by }21\text{ m}^2.\text{ If the length is increased by }1\text{ m and the breadth is decreased by }1\text{ m,}
\displaystyle \text{the area is decreased by }5\text{ m}^2.\text{ Find the perimeter of the room.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length be }l\text{ m and the breadth be }b\text{ m.}
\displaystyle (l+1)(b+1)-lb=21\qquad\ldots\text{(i)}
\displaystyle lb-(l+1)(b-1)=5\qquad\ldots\text{(ii)}
\displaystyle \text{From (i),}
\displaystyle lb+l+b+1-lb=21
\displaystyle l+b=20.
\displaystyle \text{Perimeter}=2(l+b)=2\times20=40\text{ m.}
\displaystyle \text{From (ii),}
\displaystyle lb-(lb-l+b-1)=5
\displaystyle l-b=4.
\displaystyle \text{Solving }l+b=20\text{ and }l-b=4,
\displaystyle l=12\text{ m and }b=8\text{ m.}
\displaystyle \therefore \text{The perimeter of the room is }40\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{A rectangle has twice the area of a square. The length of the rectangle is }12\text{ cm}
\displaystyle \text{greater and the width is }8\text{ cm greater than the side of the square. Find the perimeter of the square.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the square be }a\text{ cm.}
\displaystyle \therefore \text{Length of rectangle}=(a+12)\text{ cm and width}=(a+8)\text{ cm.}
\displaystyle \text{Area of rectangle}=2\times\text{Area of square}
\displaystyle (a+12)(a+8)=2a^2
\displaystyle a^2+20a+96=2a^2
\displaystyle a^2-20a-96=0
\displaystyle (a-24)(a+4)=0
\displaystyle a=24\text{ or }a=-4.
\displaystyle \text{Rejecting }a=-4,\text{ since length cannot be negative, }a=24\text{ cm.}
\displaystyle \text{Perimeter of the square}=4a=4\times24=96\text{ cm.}
\displaystyle \therefore \text{The perimeter of the square is }96\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{If the perimeter of a rectangular plot is }68\text{ m and the length of its diagonal is}
\displaystyle 26\text{ m, find its area.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length and breadth of the rectangle be }l\text{ m and }b\text{ m respectively.}
\displaystyle 2(l+b)=68
\displaystyle l+b=34.\qquad\ldots\text{(i)}
\displaystyle \sqrt{l^2+b^2}=26
\displaystyle l^2+b^2=676.\qquad\ldots\text{(ii)}
\displaystyle (l+b)^2=l^2+b^2+2lb
\displaystyle 34^2=676+2lb
\displaystyle 1156=676+2lb
\displaystyle 2lb=480
\displaystyle lb=240.
\displaystyle \therefore \text{The area of the rectangular plot is }240\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{The length of a rectangular garden is }12\text{ m more than its breadth. The numerical}
\displaystyle \text{value of its area is equal to }4\text{ times the numerical value of its perimeter. Find the dimensions}
\displaystyle \text{of the garden.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the breadth of the garden be }b\text{ m.}
\displaystyle \therefore \text{Length}=(b+12)\text{ m.}
\displaystyle \text{Area}=b(b+12).
\displaystyle \text{Perimeter}=2\{b+(b+12)\}=2(2b+12).
\displaystyle \text{Given, Area}=4\times\text{Perimeter}
\displaystyle b(b+12)=4\times2(2b+12)
\displaystyle b^2+12b=16b+96
\displaystyle b^2-4b-96=0
\displaystyle (b-12)(b+8)=0
\displaystyle b=12\text{ or }b=-8.
\displaystyle \text{Rejecting }b=-8,\text{ since breadth cannot be negative, }b=12\text{ m.}
\displaystyle \therefore \text{Length}=12+12=24\text{ m.}
\displaystyle \therefore \text{The dimensions of the garden are }24\text{ m}\times12\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{A wire when bent in the form of an equilateral triangle encloses an area of}
\displaystyle 36\sqrt{3}\text{ cm}^2.\text{ Find the area enclosed by the same wire when bent to form:}
\displaystyle \text{(i) a square, (ii) a rectangle whose length is }2\text{ cm more than its width.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the equilateral triangle be }a\text{ cm.}
\displaystyle \text{Area of an equilateral triangle}=\frac{\sqrt{3}}{4}a^2
\displaystyle \frac{\sqrt{3}}{4}a^2=36\sqrt{3}
\displaystyle a^2=144
\displaystyle a=12\text{ cm.}
\displaystyle \therefore \text{Length of the wire}=3\times12=36\text{ cm.}
\displaystyle \text{(i) When the wire is bent into a square,}
\displaystyle \text{Side of the square}=\frac{36}{4}=9\text{ cm.}
\displaystyle \text{Area of the square}=9\times9=81\text{ cm}^2.
\displaystyle \therefore \text{The area of the square is }81\text{ cm}^2.
\displaystyle \text{(ii) Let the breadth of the rectangle be }x\text{ cm.}
\displaystyle \therefore \text{Length}=(x+2)\text{ cm.}
\displaystyle 2\{x+(x+2)\}=36
\displaystyle 4x+4=36
\displaystyle 4x=32
\displaystyle x=8\text{ cm.}
\displaystyle \therefore \text{Breadth}=8\text{ cm and Length}=10\text{ cm.}
\displaystyle \text{Area of the rectangle}=10\times8=80\text{ cm}^2.
\displaystyle \therefore \text{The area of the rectangle is }80\text{ cm}^2.
\displaystyle \\

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